Question

A charged particle $q$ is placed at the centre $O$ of cube of length $L$ $(ABCDEFGH)$. Another same charge $q$ is placed at a distance $L$ from $O$. Then the electric flux through $ABCD$ is :

• A. $q / 4 \pi \varepsilon_{0} L$
• B. zero
• C. $q / 2 \pi \varepsilon_{0} L$
• D. $q / 3 \pi \varepsilon_{0} L$

Answer 1

Answer: (B)

Electric flux for any surface is defined as
$\dot{\varphi}=[\overrightarrow{ E } \cdot \overrightarrow{ d } s$

The flux through $A B C D$ can be calculated, by first taking a small elemental surface and then writing the $\overrightarrow{ E } \cdot \overrightarrow{ d }$ for this element, keep in mind that electric ficld at the location of this element is the resultant of both the chargcs. It is quite obvious the flux through $A B C D$ comes out to be non-zero because at every point of the surface, the angle between $\overrightarrow{ E }$ and $\overrightarrow{ d }$ is less than $90^{\circ}$ giving a positive non-rero value for the entire surface. So, option (b) cannot be the answer.
The options (a), (c) and (d) are dimensionally incorrect. so they cannot be answer.
Note: In this particular question flux through BCFG would be zero because component of electric field along the area vector is zero at every point on this surface.